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ELEMENTARY PROBLEMS AND SOLUTIONS
EDITED BY
RUSS EULER AND JAWAD SADEK
Please submit all new problem proposals and their solutions to the Problems Editor, DR.
RUSS EULER, Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 64468, or by email at [email protected]. All
solutions to others’ proposals must be submitted to the Solutions Editor, DR. JAWAD SADEK,
Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 64468.
If you wish to have receipt of your submission acknowledged, please include a self-addressed,
stamped envelope.
Each problem and solution should be typed on separate sheets. Solutions to problems in this
issue must be received by August 15, 2015. If a problem is not original, the proposer should
inform the Problem Editor of the history of the problem. A problem should not be submitted
elsewhere while it is under consideration for publication in this Journal. Solvers are asked to
include references rather than quoting “well-known results”.
The content of the problem sections of The Fibonacci Quarterly are all available on the web
free of charge at www.fq.math.ca/.
BASIC FORMULAS
The Fibonacci numbers Fn and the Lucas numbers Ln satisfy
Fn+2 = Fn+1 + Fn , F0 = 0, F1 = 1;
Ln+2 = Ln+1 + Ln , L0 = 2, L1 = 1.
√
√
√
Also, α = (1 + 5)/2, β = (1 − 5)/2, Fn = (αn − β n )/ 5, and Ln = αn + β n .
PROBLEMS PROPOSED IN THIS ISSUE
B-1161
Proposed by Hideyuki Ohtsuka, Saitama, Japan.
Prove each of the following:
∞
X
2k L2k
(i)
= 10,
F22k
k=1
(ii)
∞
X
k=0
tan
−1
FEBRUARY 2015
1
√
5F2k+1
=
π
.
4
81
THE FIBONACCI QUARTERLY
B-1162
Proposed by José Luis Dı́az-Barrero, Barcelona Tech, Barcelona, Spain.
Let n be a positive integer. Show that
s
n
X
p
n − 1 Fk
≤ F2n .
k−1 k
k=1
B-1163
Proposed by Ángel Plaza and Sergio Falcón, Universidad de Las Palmas
de Gran Canaria, Spain.
For any positive integer k, the k-Fibonacci and the k-Lucas sequences, {Fk,n }n∈N and
{Lk,n }n∈N , both are defined recursively by un+1 = kun + un−1 for n ≥ 1 with respective
initial conditions Fk,0 = 0, Fk,1 = 1, and Lk,0 = 2, Lk,1 = k. For any integer n ≥ 2, prove that
2
n X
kFk,j
n
(i)
≥
,
Fk,n+1 + Fk,n − 1 − kFk,j
(n − 1)2
j=1
(ii)
n X
j=1
B-1164
kLk,j
Lk,n+1 + Lk,n − 2 − k − kLk,j
2
≥
n
.
(n − 1)2
Proposed by Hideyuki Ohtsuka, Saitama, Japan.
Determine each of the following:
∞
X
(−1)n
(i)
L2Fn L2Fn+3
n=0
(ii)
∞
X
n=0
B-1165
(−1)n
.
L2Ln L2Ln+3
Proposed by Hideyuki Ohtsuka, Saitama, Japan.
For an integer n 6= 0, find the value of
LF3n−1
LF3n−2
LF3n
+
+
.
FF3n−1 FF3n−2
FF3n−2 FF3n
FF3n FF3n−1
82
VOLUME 53, NUMBER 1
ELEMENTARY PROBLEMS AND SOLUTIONS
SOLUTIONS
Easily Seen by “Telescoping”
B-1141
Proposed by Hideyuki Ohtsuka, Saitama, Japan.
(Vol. 52.1, February 2014)
Determine
∞
X
k=1
2k sin(2k θ)
.
L2k + 2 cos(2k θ)
Solution by Kenneth B. Davenport, DA, PA.
We will use the following identities:
(i) sin 2x = 2 sin x cos x.
(ii) cos 2x = 1 − 2 sin2 x = 2 cos2 x − 1.
(iii) L2n = L2n + 2(−1)n .
The key to our solution is to write the sum as a “telescoping” sum. Note that
L2k
1
1
4 cos 2k θ
−
=
.
− 2 cos 2k θ L2k + 2 cos 2k θ
L22k − 4 cos2 (2k θ)
(1)
Based on (ii) and (iii), we may write the denominator on the right-hand side as
(L2k+1 − 2 cos 2k+1 θ).
So now (1) is written:
L2k
1
1
4 cos 2k θ
=
−
.
+ 2 cos 2k θ
L2k − 2 cos 2k θ L2k+1 − 2 cos 2k+1 θ
Next, by multiplying both sides by 2k sin 2k θ and then summing we get
N
N X
X
2k sin 2k θ
2k sin 2k θ
4 · 2k sin 2k θ cos 2k θ
=
−
.
L2k + cos 2k θ
L2k − 2 cos 2k θ
L2k+1 − 2 cos 2k+1 θ
k=1
(2)
k=1
Using (i), the second term’s numerator on the right-hand side is now written 2k+1 sin 2k+1 θ.
Consequently, we now have the desired collapsing sum we were seeking, i.e.
2 sin 2θ
2N +1 sin 2N +1 θ
− N +1
.
L2 − 2 cos 2θ L2
− 2 cos 2N +1 θ
(3)
Now let N → ∞. From the Binet form of the Lucas numbers, and the fact that α2 > 2, it
is easy to see that the second term in (3) vanishes quickly. Hence, we have shown
∞
X
2k sin 2k θ
2 sin 2θ
=
.
L2k + 2 cos 2k θ
3 − 2 cos 2θ
k=1
Also solved by the proposer.
FEBRUARY 2015
83
THE FIBONACCI QUARTERLY
Summing Every Fourth Fibonacci
B-1142
Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College,
Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary
School, Buzău, Romania.
(Vol. 52.1, February 2014)
n
X
Prove that
F4k−1 = F2n · F2n+1 for any positive integer n.
k=1
Solution by Carlos Alirio Rico Acevedo, Universidad Distrital Francisco José de
Caldas (ITENU), Bogotá, Columbia.
P
2
and nk=1 Fk2 = Fn Fn+1 , see, for example [1, p.
Proof. We know that F2n+1 = Fn2 + Fn+1
79,Corollary 5.4] and [1, p. 77, Theorem 5.5], respectively. Thus,
n
X
k=1
F4k−1
n
X
2
2
=
(F2k
+ F2k−1
)
=
k=1
2n
X
Fk2
k=1
= F2n F2n+1 .
This proves the result.
References
[1] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley, New York, 2001.
All solutions were either some variations on the featured solution or they used an induction
argument.
Also solved by Brian D. Beasley, Charles K. Cook, Kenneth B. Davenport (2 solutions), Steve Edwards, Bryan Ek, Pedro Fernando Fernández Espiroza (student),
Ralph Grimaldi (2 solutions), Russell Jay Hendel, Tia Herring, Gary Knight,
Shane Latchman, Ludwing Murillo, Zibussion Ndimande, and Eyob Tarekegn
(jointly), Harris Kwong (2 solutions), Wei-kai Lai, Fidel Ngwane and Gregory Jay
(student) (jointly), Kathleen E. Lewis, Michelle Monnin, Ángel Plaza, Hideyuki
Ohtsuka, Cecil Rousseau, Jason L. Smith, Lawrence Sommer, David Stone and
John Hawkins (jointly), Dan Weiner, Nazmiye Yilmaz, and the proposer. A
nameless solution was also received.
84
VOLUME 53, NUMBER 1
ELEMENTARY PROBLEMS AND SOLUTIONS
Just Apply the AM-GM!
B-1143
Proposed by José Luis Dı́az-Barrero, Barcelona Tech, Barcelona, Spain
and Francesc Gispert Sánchez, CFIS, Barcelona Tech, Barcelona, Spain.
(Vol. 52.1, February 2014)
Let n be a positive integer. Prove that
"
#
n
n
1 X 2n Y 2
1
1−
Fk +
Fk ≥
Fn Fn+1
n
k=1
k=1
n
Y
(1−1/n)
Fk
k=1
!2
.
Solution by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain.
P
Using Fn Fn+1 = nk=1 Fk2 , the proposed inequality may be written equivalently as
!(n−1)/n n
Pn
n
n
2n
Y
Y
X
2
2
k=1 Fk
(n − 1)
+
Fk ≥
Fk
Fk2 ,
n
k=1
k=1
k=1
which follows by the AM-GM inequality.
Also solved by Dmitry Fleischman and the proposer.
It Is Not Strict When n = 1!
B-1144
Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College,
Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary
School, Buzău, Romania.
(Vol. 52.1, February 2014)
Prove that
n
Y
(Fk2 + 1) > Fn · Fn+1 + 1
(1)
(L2k + 1) > Ln · Ln+1 − 1
(2)
k=1
n
Y
k=1
for any positive integer n.
Solution 1 by Harris Kwong, SUNY Fredonia, Fredonia, NY.
Both inequalities actually become equations when n = 1, so we may assume n > 1. Then
each product contains at least two factors. Hence, since Fk2 , L2k > 0, upon expansion, we find
n
Y
(Fk2 + 1) > 1 +
k=1
and
n
Y
(L2k
k=1
FEBRUARY 2015
+ 1) > 1 +
n
X
Fk2 = 1 + Fn Fn+1 ,
k=1
n
X
k=1
L2k = 1 + (Ln Ln+1 − 2) = Ln Ln+1 − 1.
85
THE FIBONACCI QUARTERLY
Solution 2 by Sydney Marks and Leah Seader (jointly) students at California
University of Pennsylvania (CALURMA), California, PA.
Proof. We generalized inequalities (1) and (2) by proving
n
Y
(G2k + 1) ≥ Gn Gn+1 + a(a − b) + 1 for all n ∈ N,
(3)
k=1
where {Gn }n∈N is the generalized Fibonacci sequence with G1 = a and G2 = b, where a, b ∈ Z.
We first notice that
!
n
n
Y
X
2
2
(Gk + 1) ≥
Gk + 1.
(4)
k=1
k=1
We prove inequality (4) by induction on n. This inequality is clearly true when n = 1. We
assume that (4) is true for a fixed arbitrary natural number n. Thus,
!
" n
!
#
n+1
n
Y
Y
X
2
2
2
2
(Gk + 1) =
(Gk + 1) (Gn+1 + 1) ≥
Gk + 1 (G2n+1 + 1)
k=1
k=1
k=1
= G2n+1
n
X
k=1
≥
n+1
X
k=1
G2k
!
G2k
!
+
n
X
k=1
G2k
!
+ G2n+1 + 1
+ 1.
Therefore,
(4) is true for every n ∈ N. Thus, inequality (3) follows from (4) by using
Pn
2 = G G
G
n n+1 + a(a − b), see (1, Exercise 14, p. 113]. Inequalities (1) and (2) are
n=1 k
now obtained from (3) by setting a = b = 1 and a = 1, b = 3, respectively.
References
[1] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley, New York, 2001.
Also solved by Brian D. Beasley, Charles K. Cook, Kenneth B. Davenport, Steve
Edwards, Dmitry Fleischman, Ralph Grimaldi, Russell Jay Hendel, Tia Herring,
Gary Knight, Shane Latchman, Ludwig Murillo, Zibussion Ndimande and Eyob
Tarekegn (jointly) (students), Wei-kai Lai, Kathleen E. Lewis, Carolina Melee
Lopez (student), Hideyuki Ohtsuka, Ángel Plaza, David Stone and John Hawkins
(jointly), and the proposer.
Squares of “Almost Squares” Terms
B-1145
Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College,
Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary
School, Buzău, Romania.
(Vol. 52.1, February 2014)
Prove that
2 2
2
p
p
p
F1 − F1 F2 + F2 + F2 − F2 F3 + F3 + · · · + Fn − Fn F1 + F1 ≥ Fn Fn+1 (1)
86
VOLUME 53, NUMBER 1
ELEMENTARY PROBLEMS AND SOLUTIONS
2 2
2
p
p
p
L1 − L1 L2 + L2 + L2 − L2 L3 + L3 +· · ·+ Ln − Ln L1 + L1 ≥ Ln Ln+1 −2 (2)
for any positive integer n.
Solution by Ángel Plaza, Universidad de Las Palmas de Gran Canaris, Spain.
Using the AM-GM inequality, we obtain
x+y
≥
2
√
xy +
and so
q
x2 +y 2
2
2
r
,
x2 + y 2
2
for any positive real numbers x and y. Thus, the left-hand side of (1), LHS, is
x−
√
xy + y ≥
F12 + F22 F22 + F32
F 2 + F12
+
+ ··· + n
2
2
2
n
X
=
Fk2
LHS ≥
k=1
= Fn Fn+1 .
Inequality (2) is proved in the same way by using
Pn
2
k=1 Lk
= Ln Ln+1 − 2.
Also solved by Kenneth B. Davenport, Dmitry Fleischman, Russell Jay Hendel,
and the proposer.
FEBRUARY 2015
87